Problem: A few families took a trip to an amusement park together. Tickets cost $$5.50$ each for adults and $$4.50$ each for kids, and the group paid $$57.00$ in total. There were $6$ fewer adults than kids in the group. Find the number of adults and kids on the trip.
Answer: Let $x$ equal the number of adults and $y$ equal the number of kids. The system of equations is then: ${5.5x+4.5y = 57}$ ${x = y-6}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${y-6}$ for $x$ in the first equation. ${5.5}{(y-6)}{+ 4.5y = 57}$ Simplify and solve for $y$ $ 5.5y-33 + 4.5y = 57 $ $ 10y-33 = 57 $ $ 10y = 90 $ $ y = \dfrac{90}{10} $ ${y = 9}$ Now that you know ${y = 9}$ , plug it back into ${x = y-6}$ to find $x$ ${x = }{(9)}{ - 6}$ ${x = 3}$ You can also plug ${y = 9}$ into ${5.5x+4.5y = 57}$ and get the same answer for $x$ ${5.5x + 4.5}{(9)}{= 57}$ ${x = 3}$ There were $3$ adults and $9$ kids.